Download solution is a solution that contains the maximum amount of solute

CHEM 150
November Midterm
Nov. 10, 2011
B. MacLean
Name:________________________________
Instructions: The midterm consists of two sections (A, 36 points; B, 64 points). Answer each question in the
space provided. Write your name at the top of the test in the space provided. Read each question carefully.
There is a periodic table provided on the last page of the test. Including the periodic table there should be a
total of 9 pages that make up this test.
Part A: Fill-in-the-blanks. Choose the word that best completes each statement. [36 points/2 points each]
1.
A _saturated_ solution is a solution that contains the maximum amount of solute that can be dissolved
under the conditions at which the solution exists.
2.
When the chemical equation: NH3 + NO  N2 + H2O is balanced, the coefficient for the NH3 is
____4____ and the coefficient for the NO is ____6_____.
3.
The ability of an atom to attract electrons in a covalent bond is called the atom’s ___electronegativity_ .
4.
When comparing ionic and molecular compounds, it is found that ionic compounds are usually composed
of __metals_ combined with __non-metals_ while molecular compounds are usually composed of __nonmetals__ combined with ___non-metals___ .
5.
Because the vapour pressure of a solution is __lower_ than that of the pure solvent, the boiling point of
the solution will be _higher_ than that of the pure solvent.
6.
The clustering of solvent molecules around a dissolved solute molecule or ion is known as __solvation__.
7.
8.
_Electrostatic attraction*_ is an attraction or repulsion between charged species.
The process known as sublimation is a _physical_ change. During this process, heat is _absorbed_ by the
substance which sublimes.
9.
HF, hydrogen fluoride, is a polar molecule. In a liquid sample of HF, the two types of intermolecular
forces between the HF molecules are _London_ forces and _H-bonding_ forces.
10.
A bond formed between a sulfur atom and a chlorine atom will be a polar bond, with the partial negative
charge of the bond on the __chlorine_ atom.
11.
__Henry’s_ Law states that the solubility of a gas in a liquid is proportional to the partial pressure of that
gas above the liquid.
B1)
Draw Lewis structures for each of the following chemical formulas. In each case, indicate the
molecule’s (or ion’s) molecular geometry, and whether it is polar or non-polar [15 points/5 each]
i) SO32Answers:
Molecular geometry __trigonal pyramid___
Polar ? Yes /No
___________________
ii) H3O+
Molecular geometry __trigonal pyramid___
Polar ? Yes /No
___________________
ii) H2S
Molecular geometry ___angular (bent)____
Polar ? Yes /No
___________________
Notes : - don’t forget brackets in Lewis structures for ions.
A large number of students indicated that H2S was linear. Need to use VSEPR theory to predict shapes,
not Lewis structures.
B2) A sample of carbon monoxide (CO) has a volume of 12.0 L at a pressure of 730 mm Hg and a temperature
of 45oC. How many grams of CO are held in this container? [12 points]
Answer:
Have pressure (mm Hg), temperature (Celsius), volume (L) of a gas. The gas constant, R, is given in units of
(L.atm)/(mol.K). To use the PV = nRT equation:
1) Convert temperature to Kelvin: 45oC + 273 = 318 K
2) Convert pressure from mm Hg to atm:
 1 _ atm 
  0.9605263158...atm
730 _ mmHg
 760 _ mmHg 
Now use PV = nRT to solve for moles of CO:
PV  nRT
n
PV 0.9605263158...atm12.0 L 

 0.4414893553...mol _ CO
RT

L.atm 
 0.0821
318K 
mol . K 

CO has a molar mass of 28.010 g (1 mol CO = 28.010g CO), so this figure can be converted to gCO with the
following calculation:
 28.010 g _ CO 
  12.36611684...g _ CO  12.4 g _ CO
0.4414893553...mol _ CO
 1mol _ CO 
Notes:
- a lot of people blindly threw the pressure into this calculation as is (730 mm Hg) with the given gas
constant - without looking to see if the units would cancel (just a quick conversion of mm Hg to atm
using dimensional analysis).
- Three sig figs in the final answer because the Kelvin temperature has three (review significant figure
rules for addition/subtraction)
B3)
When methane (CH4) is reacted with oxygen, carbon dioxide and water are formed.
a)
Write a balanced equation describing this reaction
[5 points]
b)
How many moles of H2O are produced by combusting 10.0 g of methane?
[5 points]
Answers:
a)
CH4 + 2O2  CO2 + 2H2O
Notes: carbon dioxide has two oxygen atoms in its formula (i.e. it’s not CO)
b)
To calculate the number of moles of H2O produced, need to:
I. Convert g CH4 to mol CH4
II. Convert mol CH4 to mol H2O using conversion factor found in balanced equation
First: the molar mass of CH4 is found to be 16.0426 g (1 mol CH4 = 16.0426 g CH4)
 1mol _ CH 4  2mol _ H 2 O 

  1.246680713...mol _ H 2 O  1.25 _ mol _ H 2 O
10.0 g _ CH 4 
 16.0426 g _ CH 4  1mol _ CH 4 
Notes: a lot of people went ahead and tried to calculate the mass of water produced, rather than the number of
moles of water. Read the questions carefully.
B4
a) How many mL of water would need to be added to 25.0 mL of a 12 M NaCl solution make up a
solution having a NaCl concentration of 2.5 M?
[8 points]
Answer: a dilution problem. You want to add enough water to 25.0 mL of 12M NaCl to dilute it down to a
concentration of 2.5 M:
CsVs  CdVd
12M 25.0mL  2.5M Vd
12M 25.0mL
2.5M 
120mL  Vd
If the final volume of that solution is 120 mL, you’d need to add 95 mL (120 mL – 25.0 mL). (technically, the
answer should be 9x101 mL, as the 120 mL has the uncertain figure in the tens, but I didn’t penalize for this).
Notes: many cases where students mixed concentration/volume figures (e.g. (12M)(Vd) = (2.5M)(25.0 mL))
b) How many grams of K2SO4 are present in 25g of a solution that is 6.5%(m/m) K2SO4?
[4 points]
Answer: this is a percent-by-mass, %(m/m), calculation (formula is on back).
 mass _ of _ solute 
 x100%
Percent _ by _ mass  
 mass _ of _ solution 
 mass _ of _ solute 
 x100%
6.5%  
 25 g _ solution 
6.5%  mass _ of _ solute 


100% 
25 g

0.06525 g   mass _ of _ solute
1.625 g  mass _ of _ solute
To two sig figs, this works out to 1.6g K2SO4.
Notes: The 6.5% (m/m) means that for every 100g of solution, there are 6.5g of solute (K2SO4). If the mass of
solution isn’t 100g, but 25g (¼ of 100g), the mass of solute should be ¼ of 6.5g.
B5
a) Which of the following molecules would you expect to have the greatest dipole-dipole forces? (circle
the one, correct answer)
[3 points]
F-F
Cl-F
Br-F
I-F
Most polar molecule will have greatest dipole-dipole forces (biggest electronegativity difference)
b) For which of these molecules would you expect London forces to be greatest? (circle, as above)
[3 points]
F2
Cl2
Br2
I2
Heaviest molecule will have greatest London forces
c) Which of the following bonds would be expected to have the greatest dipole moment? (circle, as above) [3
points]
N-O
C-O
B-F
F-F
The most polar bond will have the biggest dipole moment (this is the “+” arrow drawn toward the more
polar atom of the bond)
d) Which of the following gases would you expect to be most soluble in water? (circle, as above) [3 points]
CO
CO2
N2
CH4
Most water soluble will be the only polar molecule in this list
e) Which of the following molecules should be capable of forming hydrogen bonds with each other? (circle, as
above) [3 points]
HCl
CH4
NH3
PH3
The only one of these that can H-bond is the one with either an H-O, H-N, or H-F bond